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(PARI)
A_simple(n)=local(m=2); n*=2; until(numdiv(m)==n&numdiv(m+n)==n, m++); m
A_try_pair(p, q, n, limit)=
{
/* Helper for A_prime() */
/* Look for solution which is 0 mod p^(n-1) and -n*2 mod q^(n-1) */
local(m = chinese(Mod(0, p^(n-1)), Mod(-n*2, q^(n-1))));
forstep(x=lift(m), limit, component(m, 1),
if(isprime(x\p^(n-1)) & isprime((x+n*2)\q^(n-1)), return(x)));
limit
}
A_try_above_below(m, n)=
{
/* Helper for A_prime() */
/* Function presumes that numdiv(m)==n*2 */
if(numdiv(m-n*2)==n*2, limit=m-n*2,
if(numdiv(m+n*2)==n*2, limit=m,
0))
}
A_prime(n, limit, pairmax=30)=
{
if (n%2==0 || !isprime(n), error("Only works for odd primes"));
if (default(primelimit) < limit\nextprime(pairmax+1)^(n-1),
default(primelimit, limit\nextprime(pairmax+1)^(n-1));
);
/* Evens with numdiv==n*2 are {2^(n*2-1)} u {2*p^(n-1)} u {2^(n-1)*p} */
/* Potential solutions must come from different sets */
/* Try above and below first two sets */
A_try_above_below(2^(n*2-1), n);
forprime(p=3, (limit\2)^(1/(n-1)),
if (A_try_above_below(2*p^(n-1), n), break));
/* Odd numbers with numdiv==n*2 are {p^(n*2-1)} u {p^(n-1)*q} */
/* Try where a(n) and a(n)+n*2 are (small prime)^(n-1)*(big prime) */
forprime(p=3, pairmax, forprime(q=3, pairmax,
if (p!=q, limit = A_try_pair(p, q, n, limit))));
/* Try above and below all other odd numbers with numdiv==n*2 */
forprime(p=pairmax+1, (limit\3)^(1/(n-1)),
forprime(q=3, limit\p^(n-1),
if (p!=q & A_try_above_below(p^(n-1)*q, n), break)));
forprime(p=3, limit^(1/21),
if (A_try_above_below(p^21, n), break));
limit
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